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kumpel [21]
2 years ago
12

Can someone help me on this plzzz and thank u

Mathematics
1 answer:
koban [17]2 years ago
3 0

Answer:

x = 23.4

Step-by-step explanation:

5 \cdot (x + 10) = 167\\5x + 50 = 167\\5x = 117\\x = 23.4

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<img src="https://tex.z-dn.net/?f=%28x%5E%7B2%7D%20%2Bx-3%29%3A%20%28x%5E%7B2%7D%20-4%29%5Cgeq%201" id="TexFormula1" title="(x^{
Jlenok [28]

Answer:

x>2

Step-by-step explanation:

When given the following inequality;

(x^2+x-3):(x^2-4)\geq1

Rewrite in a fractional form so that it is easier to work with. Remember, a ratio is another way of expressing a fraction where the first term is the numerator (value over the fraction) and the second is the denominator(value under the fraction);

\frac{x^2+x-3}{x^2-4}\geq1

Now bring all of the terms to one side so that the other side is just a zero, use the idea of inverse operations to achieve this:

\frac{x^2+x-3}{x^2-4}-1\geq0

Convert the (1) to have the like denominator as the other term on the left side. Keep in mind, any term over itself is equal to (1);

\frac{x^2+x-3}{x^2-4}-\frac{x^2-4}{x^2-4}\geq0

Perform the operation on the other side distribute the negative sign and combine like terms;

\frac{(x^2+x-3)-(x^2-4)}{x^2-4}\geq0\\\\\frac{x^2+x-3-x^2+4}{x^2-4}\geq0\\\\\frac{x+1}{x^2-4}\geq0

Factor the equation so that one can find the intervales where the inequality is true;

\frac{x+1}{(x-2)(x+2)}\geq0

Solve to find the intervales when the equation is true. These intervales are the spaces between the zeros. The zeros of the inequality can be found using the zero product property (which states that any number times zero equals zero), these zeros are as follows;

-1, 2, -2

Therefore the intervales are the following, remember, the denominator cannot be zero, therefore some zeros are not included in the domain

x\leq-2\\-2

Substitute a value in these intervales to find out if the inequality is positive or negative, if it is positive then the interval is a solution, if it is negative then it is not a solution. This is because the inequality is greater than or equal to zero;

x\leq-2   -> negative

-2   -> neagtive

-1\leq x   -> neagtive

x>2   -> positive

Therefore, the solution to the inequality is the following;

x>2

6 0
2 years ago
pls help look at attached image choose any problem u would like to do this is for my middle school math class again pls help and
Sonbull [250]
Can you take a picture again I can’t see anything
3 0
11 months ago
Read 2 more answers
consider the quadratic form q(x,y,z)=11x^2-16xy-y^2+8xz-4yz-4z^2. Find an orthogonal change of variable that eliminates the cros
Bezzdna [24]

Answer:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

Step-by-step explanation:

The given quadratic form is of the form

q(x,y,z)=ax^2+by^2+dxy+exz+fyz.

Where a=11,b=-1,c=-4,d=-16,e=8,f=-4.Every quadratic form of this kind can be written as

q(x,y,z)={\bf x}^{T}A{\bf x}=ax^2+by^2+cz^2+dxy+exz+fyz=\left(\begin{array}{ccc}x&y&z\end{array}\right) \left(\begin{array}{ccc}a&\frac{1}{2} d&\frac{1}{2} e\\\frac{1}{2} d&b&\frac{1}{2} f\\\frac{1}{2} e&\frac{1}{2} f&c\end{array}\right) \left(\begin{array}{c}x&y&z\end{array}\right)

Observe that A is a symmetric matrix. So A is orthogonally diagonalizable, that is to say,  D=Q^{T}AQ where Q is an orthogonal matrix and D is a diagonal matrix.

In our case we have:

A=\left(\begin{array}{ccc}11&(\frac{1}{2})(-16) &(\frac{1}{2}) (8)\\(\frac{1}{2}) (-16)&(-1)&(\frac{1}{2}) (-4)\\(\frac{1}{2}) (8)&(\frac{1}{2}) (-4)&(-4)\end{array}\right)=\left(\begin{array}{ccc}11&-8 &4\\-8&-1&-2\\4&-2&-4\end{array}\right)

The eigenvalues of A are \lambda_{1}=16,\lambda_{2}=-5,\lambda_{3}=-5.

Every symmetric matriz is orthogonally diagonalizable. Applying the process of diagonalization by an orthogonal matrix we have that:

Q=\left(\begin{array}{ccc}\frac{4}{\sqrt{21}}&-\frac{1}{\sqrt{17}}&\frac{8}{\sqrt{357}}\\\frac{-2}{\sqrt{21}}&0&\sqrt{\frac{17}{21}}\\\frac{1}{\sqrt{21}}&\frac{4}{\sqrt{17}}&\frac{2}{\sqrt{357}}\end{array}\right)

D=\left(\begin{array}{ccc}16&0&0\\0&-5&0\\0&0&-5\end{array}\right)

Now, we have to do the change of variables {\bf x}=Q{\bf y} to obtain

q({\bf x})={\bf x}^{T}A{\bf x}=(Q{\bf y})^{T}AQ{\bf y}={\bf y}^{T}Q^{T}AQ{\bf y}={\bf y}^{T}D{\bf y}=\lambda_{1}y_{1}^{2}+\lambda_{2}y_{2}^{2}+\lambda_{3}y_{3}^{2}=16y_{1}^{2}-5y_{2}^{2}-5y_{3}^2

Which can be written as:

q(x,y,z)=16x^{2}-5y^{2}-5z^{2}

4 0
3 years ago
If 7x + 4 = -19 + 5x, then 2x – 14 equals A) 23 B) -23 C) -3 D) 16 E) NOTA
JulsSmile [24]

Answer:

2x-14 = -37

Step-by-step explanation:

7x + 4 = -19 + 5x,

Subtract 5x from each side

7x-5x + 4 = -19 + 5x -5x

2x +4 = -19

Subtract 4 from each side

2x +4-4 = -19 -4

2x = -23

We want to find 2x -14

2x = -23

2x-14

-23-14

-37

7 0
3 years ago
Read 2 more answers
How do you model a decimal using a fraction
Darya [45]
Move over two places
5 0
3 years ago
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